Problem of the Day 6

$f(0)=2$ is the global maximum; this global maximum is attained only when $\cos(x)=\cos(\sqrt{2}x)=1.$ Now $\cos(x)=1$ when $x=2\pi{k}$ for an integer $k$ , and $\cos(\sqrt{2}x)=1$ when $x=\sqrt{2}\pi{k}$ . For nonzero $k_1$ and $k_2$ , we have $2\pi{k_1}\neq\sqrt{2}\pi{k_2}$ since $\sqrt{2}$ is irrational, so that the global maximum 2 is attained at $x=0$ only.